A rational function is a fraction where both the numerator and denominator are polynomials. When analyzing and/or graphing rational functions the zeros of the two polynomials tell you most of what you need to know. And since zeros of polynomials are found by factoring them, factoring is a huge part of analyzing and/or graphing rational functions.
So we start by factoring the numerator and denominator. The denominator is a trinomial that factors very easily:
The numerator will factor by grouping:
We can continue to factor the numerator using the difference of squares pattern:
The function is now fully factored. And in factored form we can see (or figure out) the values of x that make a factor zero. And if a factor is zero, the product is zero.
The values of x that make the numerator zero are -2, -1 and 1. The values of x that make the denominator zero are -3 and 2.
If you can't "see" these numbers in the factors, then set each factor equal to zero and solve the equation. For example, for the factor x+2:
x+2 = 0
Subtract 2 from each side:
x = -2
Now we're ready to figure out the details of your function:- "Holes" (I think the technical term is "removable discontinuity") occur for x values that make both the numerator and denominator zero. (SPECIAL CASE: If every factor of the denominator represents a "hole" then your function is a special case. An explanation of how to handle this special case is at the end.) There are no "holes" in your function.
- Vertical asymptotes occur for x values that make just the denominator zero. So your function will have vertical asymptotes at x = -3 and x = 2.
- X-intercepts occur for x values that make just the numerator zero. So your function will have x-intercepts at -2, -1 and 1.
For the rest of what we need, we'll use the original, non-factored form of the function, :- The y-intercept occurs when x = 0. This can be found easily because if x is zero any term with x in it will be zero, too. Looking at your function we should be able to see that if x is zero, the numerator becomes -2 and the denominator becomes -6 making the function have a value of -2/-6 or 1/3 when x is zero. So the y-intercept is 1/3.
- Horizontal, oblique and non-linear asymptotes. These are determined in large part by the degrees (highest exponents) of the numerator and denominator:
- If the degree of the numerator is less than the degree of the denominator, then the line y = 0 will be a horizontal asymptote.
- If the degrees are equal, then there will be a horizontal asymptote at the value of the ratio of leading coefficients (the coefficients of the terms with the highest exponent). For example, the function has leading coefficients of 3 and 2. Its horizontal asymptote will be y = 3/2.
- If the degree of the numerator is greater than the degree of the denominator, then there will not be a horizontal asymptote. But there will be either an oblique or a non-linear asymptote. (These are not always taught so you may not be responsible for finding them.) The degree of your numerator is 3 and the degree of your denominator is 2. So you will have a either an oblique or a non-linear asymptote.
The logic behind all of this is based on what happens to fractions when x becomes very large positive or negative numbers.
Finding an oblique or a non-linear asymptote:- Divide the fraction.
- Discard the remainder. (If there is no remainder, then you missed the SPECIAL CASE mentioned above under "Holes" and explained below.)
- The remaining equation is an asymptote. If it is the equation of a line, then it is called an oblique asymptote. Otherwise it is a non-linear asymptote.
Since your function has this type of asymptote we will work through these steps on your function:
1. Divide
x + 1
________________________________
x^2+x-6 /x^3 +2x^2 -x -2
x^3 + x^2 -6x
----------------
x^2 +5x -2
x^2 + x -6
------------
4x +4
So
2. Discard the remainder.
(Since this changes the equation I will no longer call it f(x)):
y = x + 1
3. So y = x + 1 is an asymptote. This is the equation of a line so it is called an oblique asymptote.
Here's a look at the graph of your function. (The obliques asymptote is in green. The vertical asymptotes, x = -3 and x = 2, are not drawn because I do not know how to get Algebra.com's software to draw them. I hope they are obvious.):
**SPECIAL CASE. If every factor of the denominator represents a "hole":- Cancel the common factors in the numerator and denominator.
- You now have a function which is not a rational function. The graph of the original function will be the same as the graph of this new, non-rational function except there will be be "holes" in it for the x values that make the original denominator zero.
For example, if then, after canceling, we have just x+7. y = x+7 is a line. The graph of g(x) will be this same line except that there will be "holes" at x = 3 and x = -4.